The normal at the point (1,1) on the curve 2y | vedclass

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(B) The equation of the given curve is $2y + x^{2} = 3$.Differentiating with respect to $x$,we get:$2 \frac{dy}{dx} + 2x = 0$$\frac{dy}{dx} = -x$At th

Vedclass · Accepted Answer

$x-y=0$

Vedclass · Answer

(B) The equation of the given curve is $2y + x^{2} = 3$.Differentiating with respect to $x$,we get:$2 \frac{dy}{dx} + 2x = 0$$\frac{dy}{dx} = -x$At the point $(1,1)$,the slope of the tangent is $\left. \frac{dy}{dx} ight|_{(1,1)} = -1$.The slope of the normal at $(1,1)$ is given by $m_{normal} = \frac{-1}{	ext{slope of tangent}} = \frac{-1}{-1} = 1$.The equation of the normal at $(1,1)$ is $y - y_{1} = m_{normal}(x - x_{1})$.$y - 1 = 1(x - 1)$$y - 1 = x - 1$$x - y = 0$.Thus,the correct option is $B$.

The normal at the point $(1,1)$ on the curve $2y + x^{2} = 3$ is

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